Question

A(n) 60.3 g ball is dropped from a height of 53.7 cm above a spring of negligible mass. The ball compresses the spring to a maximum displacement of 4.68317 cm. The acceleration of gravity is 9.8 m/s 2 . x h Calculate the spring force constant k

Answer 1

Answer:

271.862 N/m

Explanation:

From Hook's Law,

mgh = 1/2ke²............... Equation 1

Where

m = mass of the ball, g = acceleration due to gravity, k = spring constant, e = extension, h = height fro which the ball was dropped.

Making k the subject of the equation,

k =2mgh/k²....................... Equation 2

Note: The potential energy of the ball is equal to the elastic potential energy of the spring.

Given: m = 60.3 g = 0.0603 kg, g = 9.8 m/s², e = 4.68317 cm = 0.0468317 m, h = 53.7 cm = 0.537 m

Substitute into equation 2

k = 2(0.0603)(9.8)(0.537)/0.048317²

k = 0.6346696/0.0023345

k = 271.862 N/m